python - 如何将矩形矩阵转换为随机且不可约矩阵?
问题描述
我编写了以下代码来将矩阵转换为随机且不可约矩阵。我遵循了一篇论文(Deeper Inside PageRank)来编写这段代码。此代码适用于方阵,但会给出矩形矩阵的错误。如何修改它以将矩形矩阵转换为随机和不可约矩阵?
我的代码:
import numpy as np
P = np.array([[0, 1/2, 1/2, 0, 0, 0], [0, 0, 0, 0, 0, 0], [1/3, 1/3, 0, 0, 1/3, 0], [0, 0, 0, 0, 1/2, 1/2], [0, 0, 0, 1/2, 0, 1/2]])
#P is the original matrix containing 0 rows
col_len = len(P[0])
row_len = len(P)
eT = np.ones(shape=(1, col_len)) # Row vector of ones to replace row of zeros
e = eT.transpose() # it is a column vector e
eT_n = np.array(eT / col_len) # obtained by dividing row vector of ones by order of matrix
Rsum = 0
for i in range(row_len):
for j in range(col_len):
Rsum = Rsum + P[i][j]
if Rsum == 0:
P[i] = eT_n
Rsum = 0
P_bar = P.astype(float) #P_bar is stochastic matrix obtained by replacing row of ones by eT_n in P
alpha = 0.85
P_dbar = alpha * P_bar + (1 - alpha) * e * (eT_n) #P_dbar is the irreducible matrix
print("The stocastic and irreducible matrix P_dbar is:\n", P_dbar)
预期输出:
A rectangular stochastic and irreducible matrix.
实际输出:
Traceback (most recent call last):
File "C:/Users/admin/PycharmProjects/Recommender/StochasticMatrix_11Aug19_BSK_v3.py", line 13, in <module>
P_dbar = alpha * P_bar + (1 - alpha) * e * (eT_n) #P_dbar is the irreducible matrix
ValueError: operands could not be broadcast together with shapes (5,6) (6,6)
解决方案
您正在尝试将两个不同形状的数组相乘。这是行不通的,因为一个数组有 30 个元素,而另一个数组有 36 个元素。
您必须确保该数组e * eT_n
与您的输入数组具有相同的形状P
。
您没有使用该row_len
值。但如果e
行数正确,您的代码将运行。
# e = eT.transpose() # this will only work when the input array is square
e = np.ones(shape=(row_len, 1)) # this also works with a rectangular P
您可以检查形状是否正确:
(e * eT_n).shape == P.shape
您应该学习 numpy 文档和教程以了解如何使用 ndarray 数据结构。它非常强大,但也与原生 python 数据类型有很大不同。
例如,您可以用向量化数组操作替换这个冗长且非常慢的嵌套 python 循环。
原始代码(带有固定缩进):
for i in range(row_len):
Rsum = 0
for j in range(col_len):
Rsum = Rsum + P[i][j]
if Rsum == 0:
P[i] = eT_n
惯用的 numpy 代码:
P[P.sum(axis=1) == 0] = eT_n
此外,您不需要创建数组eT_n
。由于它只是重复的单个值,因此您可以直接分配标量 1/6。
# eT = np.ones(shape=(1, col_len))
# eT_n = np.array(eT / col_len)
P[P.sum(axis=1) == 0] = 1 / P.shape[1]
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